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Maksim231197 [3]
4 years ago
12

For your senior project, you would like to build a cyclotron that will accelerate protons to 10% of the speed of light. The larg

est vacuum chamber you can find is 60 cm in diameter.
Physics
1 answer:
DIA [1.3K]4 years ago
7 0

The given question is incomplete. The complete question is as follows.

For your senior project, you would like to build a cyclotron that will accelerate protons to 10% of the speed of light. The largest vacuum chamber you can find is 60 cm in diameter.

What magnetic field strength will you need?

Explanation:

Formula for the strength of magnetic field is as follows.

      B = \frac{mv}{qr}

Here,    m = mass of proton = 1.67 \times 10^{-27} kg

      v = velocity = 10% of 3 \times 10^{8} = 3 \times 10^{7} m/s

      q = charge of proton = 1.6 \times 10^{-19} C

      r = radius = \frac{60}{2} = 30 cm = 0.30 m   (as 1 m = 100 cm)

Therefore, magnetic field will be calculated as follows.

            B = \frac{mv}{qr}

               = \frac{1.67 \times 10^{-27} \times 3 \times 10^{7}}{1.6 \times 10^{-19} C \times 0.30 m}

               = \frac{5.01 \times 10^{-20}}{0.48 \times 10^{-19}}

               = 1.0437 T

Thus, we can conclude that magnetic field strength is 1.0437 T.

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7. Nancy has a mass of 60 kg and sits on the very end of a 3.00 m long plank pivoted in the middle. How much torque must her co-
netineya [11]

Answer:

Torque = 882Nm

Explanation:

Torque = Mg×distance

But plank's is pivoted ,therefore distance=3/2=1.5m

Mass of Nancy=60jg

Acceleration due to gravity, g=9.8m/s^2

Torque= 60×9.8×1.5

Torque= 882Nm

6 0
3 years ago
A calorimeter is a container that is insulated from the outside, so a negligible amount of energy enters or leaves the container
Len [333]

Answer:

This is a heat balance question with negligible heat losses  to the surroundings.

Sum of heat losses = Sum of the heat gains

We are required to find the final temperature of the calorimeter system T1?

we were given

C = specific heat capacities of Copper c, Water w, Ice i and Lead l

m = masses as mc = 0.1kg, mw = 0.16kg, mi=0.018kg, ml=0.75kg,

T = Temperature of the components, Lead Tl= 225C and at ambient pressure, Temp. To of the water, ice and copper = 25C

Since, at thermal equilibrium after adding the Lead, the lead temperature Tl decreases while copper, water and ice temp decrease, so we have respective heat losses 'q' as,

qc + qw + qi + ql =0

which means that

(mc x Cc x ΔTC) + (mw x Cw x ΔTC) + (mi x Ci x ΔTC) + (ml x Cl x ΔTC) = 0

(0.1x390x (T1 - 25)) + (0.16x4190x (T1 - 25)) + (0.018x4190x (T1 - 0)) + (0.750x130x (T1 - 225)) = 0

882.32T1 - 41558 = 0

T1 = 47.1K

Hence, the final temperature of the calorimeter system is 47.1K

3 0
4 years ago
A tennis player practices against a wall, hitting a 0.1 kg ball towards the
pantera1 [17]

Answer: 80 Newton

Explanation:

Initial velocity of ball = +20 m/s.

Final velocity of ball = -20 m/s

Mass of ball = 0.1kg

Time taken = 0.05 seconds

Average force = (Change in momentum of moving ball / Time taken)

Since, change in momentum = Mass (final velocity - initial velocity)

Change in momentum =0.1 x (-20 - (+20))

= 0.1 x (-20-20)

= 0.1 x (-40)

= -4.0 kgm/s

Then, put -4.0 kgm/s in the equation of force when Average Force = (Change in momentum / Time taken)

= (-4.0kgm/s / 0.05 seconds)

= 80Newton (note that the negative sign does not reflect on the magnitude of force)

Thus, the average force exerted on the ball is 80N

3 0
3 years ago
Give an example of an energy conversion that produces an enwanted from of energy
stira [4]
An example of an energy conversion that produces an unwanted form of energy is mixing acids with water. 
4 0
3 years ago
A particle moves along the x-axis according to x(t)=10t−2t²m. (a) What is the instantaneous velocity at t = 2 s and t = 3 s? (b)
Alisiya [41]

Answer:

a) v(2) = 2m/s, v(3) = -2m/s

b) speed at t = 2s is 2m/s

speed at t = 3s is 2m/s

c) 0 m/s

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We can take the derivative of x(t) to find the equation of velocity

v(t) = x'(t) = 10 - 4t

(a) v(2) = 10 - 4*2 = 10 - 8 = 2 m/s

v(3) = 10 - 4*3 = 10 - 12 = -2 m/s

(b) The speed would be the same as velocity without the direction

speed at t = 2s is 2m/s

speed at t = 3s is 2m/s

(c) The average velocity between t = 2s and t = 3s is distance it travels over period of time

v_a = \frac{s(3) - s(2)}{\Delta t} = \frac{10*3 - 2*3^2 - (10*2 - 2*2^2)}{3 - 2}

v_a = \frac{12 - 12}{1} = 0/1 = 0 m/s

6 0
3 years ago
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